$\iint_{D} e^{\sin x \cos y} dA$

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I have a double integral

$$\iint_{D} e^{\sin x \cos y} dA$$

where $D$ is the disk with center the origin and radius $3$.

My guess would be to estimate the integral by a change of variables technique, but I can't seem to know where to start from. Maybe I should perform a trigonometric substitution?

Any hints and solutions would be highly appreciated.

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Using one of the properties of double integrals

If $ {m \le f(x,y) \le M} $ for all $ (x,y) $ in $D$, then $$ mA(D) \le \iint_D f(x,y)dA \le MA(D) $$

We could estimate the given integral $ \iint_D e^{\sin x\ \cos y} dA$, where $D$ is the circle with radius $3$

As Benjamin Moss noted above, $-1 \leq \sin(x)\cos(y)\leq 1$

Now we can proceed to $$ e^{-1} \le e^{\sin x\ \cos y} \leq e ,$$

where $m=e^{-1}$ and $M=e$.

$A(D)$ which is the area of the circle has a formula of $A=\pi r^2$

Therefore, our final estimation would be $$ {(9\pi)\over {e}} \le \iint_D e^{\sin x \cos y} dA \le 9\pi e$$